Question
Factorise : $(2a + b)^2- 6a - 3b - 4$
✓
Answer
$(2a + b)^2- 6a - 3b - 4$
$= ( 2a + b )^2 - 3( 2a + b ) - 4$
Assume that, $2a + b = x$
Therefore,
$(2a + b)^2- 6a - 3b - 4$
$= x2 - 3x - 4$
$= x2 - 4x + x - 4$
$= 1( x - 4 ) + x( x - 4 )$
$= ( x + 1 )( x - 4 )$
$= ( 2a + b + 1 )( 2a + b - 4 )$
$[$ resubstitute the value of $x ]$
$= ( 2a + b )^2 - 3( 2a + b ) - 4$
Assume that, $2a + b = x$
Therefore,
$(2a + b)^2- 6a - 3b - 4$
$= x2 - 3x - 4$
$= x2 - 4x + x - 4$
$= 1( x - 4 ) + x( x - 4 )$
$= ( x + 1 )( x - 4 )$
$= ( 2a + b + 1 )( 2a + b - 4 )$
$[$ resubstitute the value of $x ]$
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